Mechanical energy question

[Tokspor]

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I'm confused about this issue. Let's say there are two objects of equal masses starting at the same height. One is on an inclined plane and the other is just in the air. I understand that if they are both allowed to fall, their final velocities should be the same because mgh = (1/2) mv^2.

But I'm wondering, what about the time it takes for them to fall as well as the work it takes? For the object on the inclined plane, the acceleration would only be some fraction of g, gsin(theta). So wouldn't it take longer to get to the final point? Also, since work is a path function, shouldn't the object on the inclined plane also have to exert more work?

 

[TheBoondocks]

I'm confused about this issue. Let's say there are two objects of equal masses starting at the same height. One is on an inclined plane and the other is just in the air. I understand that if they are both allowed to fall, their final velocities should be the same because mgh = (1/2) mv^2.

But I'm wondering, what about the time it takes for them to fall as well as the work it takes? For the object on the inclined plane, the acceleration would only be some fraction of g, gsin(theta). So wouldn't it take longer to get to the final point? Also, since work is a path function, shouldn't the object on the inclined plane also have to exert more work?

Here's something 99% of premeds screw up. Net work Fd= change in KE
Now, if I lift a ball to a shelf and set it there, I did NO net work. I did positive work of mgh BUT gravity did negative work. So, the NET WORK is zero.

In your case, if you drop the ball, the final velocity is (2gh)^.5 and the KE equals the work done.

In the case of the incline, the final velocity is the same and the inital height was the same, so the net work must be the same.

What you're thinking of is POWER which is the rate of work. The acceleration of the incline is gsine theta. If you drop the ball it will reach the ground faster than the incline. However, assuming no friction, the rolling ball will reach the end of the incline with the same velocity with the same work.

With work be careful with respect to whether it is net work. If it is, then always you KE/ ME.

Work is a path function when you are considering NON net work. If I push a cart with a force of 10 Newtons for 10 meters and you do the same except for 100 meters. You have done more WORK than I did. However, our net work is 0 because the KE is 0.

Finally, if you accelerate a box from 0 to 10m/s over 100 meters and I do it over 10 meters, then both of us have done the same NET WORK because the KE is the same. Assuming constant acceleration, the initial velocity is 0 and final is 10. vo is 0 and vf is 10. So 100=2a(10) So MY acceleration is 5.
You on the otherhand, your acceleration is 100=200= .5m/s^2.

So, path matters that is distance, but different force can be the great equalizer.

 

[RogueUnicorn]

Here's something 99% of premeds screw up. Net work Fd= change in KE
Now, if I lift a ball to a shelf and set it there, I did NO net work. I did positive work of mgh BUT gravity did negative work. So, the NET WORK is zero.

no, you have done positive work. gravity did negative work. the NET work is zero.

 

[ezsanche]

The object that fell straight down takes the shortest path to the ground. and has the greatest acceleration. The object that fell at a downward slant takes a longer path to the ground and accelerate slower than the first object.

 

[ezsanche]

1

 

[TheBoondocks]

no, you have done positive work. gravity did negative work. the NET work is zero.

I am correct. It is possible to do positive work without doing net work. If I move a box 10 meters with a force of 10 Newtons then I did 100 Joules of work. However, the net work was 0 assuming it started at 0 and ended at 0. If I throw a ball and accelerate it over a distance of 1 meter, gravity does negative work and I do positive work. In this case my positive work outweighs the negative gravity so I do net work.

So, in the example above, I don't do net work (on the box which is what I was implying) in moving the box is because of the lack of change in KE. I do positive work. However, I do no net work on the box. That is correct. The reason I don't do net work (change in KE) is because my positive work is counteracted by gravity and 0 KE change.. I even noted the net work is 0 in my original post. It is more accurate to say the net work is 0 when going verticle. However, when moving horizontal, it is not inaccurate to say a person did no net work or in lifting a box upwards you do no net work. One generally assumes when I say I don't do net work, it means that in lifting the box the net work on it was 0. Of course I do positive work. If this was misconstrued, I later clearly say

Here's something 99% of premeds screw up. Net work Fd= change in KE
Now, if I lift a ball to a shelf and set it there, I did NO net work. I did positive work of mgh BUT gravity did negative work. So, the NET WORK is zero.

just like you said so it should have been surmised that I'm aware of the positive/negative dynamic. I will find a link. However, a physic's professor explains this topic and says YOU do no NET WORK in moving a box up the stairs and I used similar terminology. I see your point in being more specific, which I later stated in my sentence. However, even if I hadn't stated exactly what you said showing my comprehension, it was still acceptable use of terms.

 

[RogueUnicorn]

relax. i know what you're saying. but saying "i do net work of zero" is potentially confusing.

 

[TheBoondocks]

relax. i know what you're saying. but saying "i do net work of zero" is potentially confusing.

cool. I agree with the potentially confusing part. I took a Kaplan test that tested this very concept. You're right about those topicals. They're hard, but really force you to think. I love the Kaplan BS tests, I don't see how test day can be any harder. The BS is ALL reading comprehension. So, did you take all tests 1-11. I'm wondering if it's worth doing 10 and 11?

 

[RogueUnicorn]

cool. I agree with the potentially confusing part. I took a Kaplan test that tested this very concept. You're right about those topicals. They're hard, but really force you to think. I love the Kaplan BS tests, I don't see how test day can be any harder. The BS is ALL reading comprehension. So, did you take all tests 1-11. I'm wondering if it's worth doing 10 and 11?

honestly the kaplan FLs after 5 or 6 are kind of crap IMO. better to save your time and do the AAMCs (which you should do anyway). the topicals really are the most indispensable part of the online stuff.

 

[BennieBlanco]

relax. i know what you're saying. but saying "i do net work of zero" is potentially confusing.

agreed

 

[BennieBlanco]

honestly the kaplan FLs after 5 or 6 are kind of crap IMO. better to save your time and do the AAMCs (which you should do anyway). the topicals really are the most indispensable part of the online stuff.

thx, I was not aware of this.

 

[TieuBachHo]

Nevermind.

 

[TheBoondocks]

To the OP,
This is just from what I can recall / my .02 cents.

1) For conservative forces, the work is indepedent of paths taken. Use W = Fd cos(theta) to calculate.
2) The time it takes and the accelerations are the same. However, you have both X and Y components for the inclined plane and it might allude you from thinking that they are not since a(x) = g sin(theta). You also need a Y component for the inclined acceleration.
3) Now, if you are still confused then you can relate this to a projectile motion.

In the end, just work it out like any other problems until you get a grasp of this concept.

The accelerations aren't the same. The ball falls with an acceleration of g. The projectile is different because the ball thrown horizontally still has an acceleration of g just like the ball that was dropped vertically in the y direction. This isn't the case comparing the incline to the vertical drop. In the incline if it's sliding, it's vertical acceleration is not g it's gsine(theta.) Then of that agsine(theta), you have x and y and the y obviously is less than g.

 

[TheBoondocks]

The object that fell straight down takes the shortest path to the ground. and has the greatest acceleration. The object that fell at a downward slant takes a longer path to the ground and accelerate slower than the first object.

Exactly.

 

[TieuBachHo]

The accelerations aren't the same. The ball falls with an acceleration of g. The projectile is different because the ball thrown horizontally still has an acceleration of g just like the ball that was dropped vertically in the y direction. This isn't the case comparing the incline to the vertical drop. In the incline if it's sliding, it's vertical acceleration is not g it's gsine(theta.) Then of that agsine(theta), you have x and y and the y obviously is less than g.

You are right. It's my mistake. I need to do more review on this. Thanks!

 

[TheBoondocks]

You are right. It's my mistake. I need to do more review on this. Thanks!

No that's good. Tha'ts how you kill the MCAT. Really think about the concepts. The MCAT isn't hard. I realized I didn't understand stuff as well as I thought. When you read a question, always make sure you comprehend what it's asking. To illustrate I will give the following example.

We have an acetic acid solution which is obviously acidic because it is a weak acid. acetic acid(CH3COOH) and it's conjugate base acetate anion (CH3COO-) can act as a buffer.

The question asked the following, what will happen to the pH of the solution upon adding NaCH3COO a salt. Well, I FOOLISHLY answered the pH would decrease (meaning get more acidic) because it's a weak acid.

That's not what the question asked!!! Na is irrelevant. CH3COO- is a weak base. What happens when you add a base to anything? pH goes UP. So, make sure you're thinking correctly. With practice this will become second nature and you'll kill it. Bleargh who posted earlier in this thread got a 15 PS 11 Vr and 15 BS. He said that he was getting 12s and wasn't missing anything because of knowledge. He made a conscientious effort to eliminate stupid mistakes.

Finally, don't call a mistake stupid. A math error due to say converting 1.6 e-11 to 16 e -10 when it should be 16 e-12. Any other error was made because you made an assumption. Just read carefully and then use your knowledge to apply.